Answer
$Area=\dfrac{1}{2}$
Work Step by Step
Here, we have: $y=x^3, y=1-x^3, x=0$ to $x=1$
The area is given by $A=\int_0^{1} [(1-x^3)-x^3] \ dx=\int_0^1 (1-2x^3) \ dx$
or, $= [x-\dfrac{2x^4}{4}]_0^1$
or, $=[x-\dfrac{x^4}{2}]_0^1$
or, $=[1-\dfrac{(1)^4}{2}]-[0-\dfrac{(0)^4}{2}]$
Therefore, the required area is: $Area=\dfrac{1}{2}$
